What Are the Tensions in the Strings?

[Masrat_A]

Homework Statement

In Figure 5, the two strings attached to the mass are fixed to the ceiling and the wall, respectively. Determine the tensions in the strings.

Figure: http://i.imgur.com/jbTo1Nnh.jpg

Homework Equations

Please look below.

The Attempt at a Solution

Ceiling $= 37^o$
Wall $= 137^o$
Object $= 5 kg$

$F_t = F_g$
$F_t = Mg$
$F_t = 5 * 10$
$F_t = 50N$

Ceiling
$F_1 = 50cos37^o$
$F_1 = 40N$

Wall
$F_1 = 50sin137^o$
$F_1 = 34N$

Does this look reasonable?

 

[andrewkirk]

There are a number of problems:

• The diagram says the angle between the wall string and the horizontal is 137 degrees, which is obtuse, yet it is an acute angle.
• The solution uses the symbol $F_1$, which is undefined.

In the free body diagram, using $T_c$ and $T_w$ for the tension in the ceiling and wall strings respectively, and $T_{c,x},T_{w,x}$ for the horizontal components and $T_{c,y},T_{w,y}$ for the vertical components, you need to write out formulas for each of those four components in terms of $T_c,T_w$ and the two angles.
Then write two equations, one to require the total of all vertical ($y$) components to be zero, and the other for the horizontal ($x$) components.

You will have two unknowns: $T_c,T_w$ and two equations, so you can solve them.

 

[Masrat_A]

The diagram says the angle between the wall string and the horizontal is 137 degrees, which is obtuse, yet it is an acute angle.

I don't have much trigonometry background, unfortunately; could you please expand on this a little more? Does this mean I cannot be using sine, and if not, what would you recommend me to do?

The solution uses the symbol $F_1$, which is undefined.

In the free body diagram, using $T_c$ and $T_w$ for the tension in the ceiling and wall strings respectively, and $T_{c,x},T_{w,x}$ for the horizontal components and $T_{c,y},T_{w,y}$ for the vertical components, you need to write out formulas for each of those four components in terms of $T_c,T_w$ and the two angles.
Then write two equations, one to require the total of all vertical ($y$) components to be zero, and the other for the horizontal ($x$) components.

You will have two unknowns: $T_c,T_w$ and two equations, so you can solve them.

Could you please explain to me what the formulas for x and y would be? Afterward, in what way shall I set the vertical and horizontal components to zero?

 

[TomHart]

I don't have much trigonometry background, unfortunately; could you please expand on this a little more? Does this mean I cannot be using sine, and if not, what would you recommend me to do?

From the figure, it looks to me like the angle is 37 degrees, not 137 degrees. I think what is being interpreted as a "1" is actually the angle indicator - the same as is the case in Figure 1.

 

[Masrat_A]

From the figure, it looks to me like the angle is 37 degrees, not 137 degrees. I think what is being interpreted as a "1" is actually the angle indicator - the same as is the case in Figure 1.

Would this then allow me to use cosine on the angle?

 

[Nidum]

Yes . . Both angles seem to be the same at 37 degrees .

 

[Masrat_A]

In the free body diagram, using $T_c$ and $T_w$ for the tension in the ceiling and wall strings respectively, and $T_{c,x},T_{w,x}$ for the horizontal components and $T_{c,y},T_{w,y}$ for the vertical components, you need to write out formulas for each of those four components in terms of $T_c,T_w$ and the two angles.

Would these be the vertical and horizontal components for $T_c$ and $T_w$?

$T_{c,x} = 50cos{37} = 39.93$
$T_{c,y} = 50cos{37} = 30.09$

$T_{w,x} = 50cos{37} = 39.93$
$T_{w,y} = 50cos{37} = 30.09$

Then write two equations, one to require the total of all vertical ($y$) components to be zero, and the other for the horizontal ($x$) components. You will have two unknowns: $T_c,T_w$ and two equations, so you can solve them.

I'm still a little confused on this however. Could you please expand on it a little more?

 

[haruspex]

Would these be the vertical and horizontal components for $T_c$ and $T_w$?

$T_{c,x} = 50cos{37} = 39.93$
$T_{c,y} = 50cos{37} = 30.09$

$T_{w,x} = 50cos{37} = 39.93$
$T_{w,y} = 50cos{37} = 30.09$
?

No. By what law of physics do you get those equations?

If the overall tension in the ceiling attachment is T_c, what is its horizontal component?
What horizontal forces act on the mass? For equilibrium, what equation can you write?

 

[Masrat_A]

I'm sorry; I misinterpreted what was written on my textbook. Upon further inspection, here is what I have ended up with so far. Do any of these seem right?

For the string attached to the wall, our degree is 37; there is a 90 degree angle formed by the dotted lines in our figure above, and 37 subtracted from 90 gives us 53.

$T_1cos37^o = T_2cos53^o$
$T_1sin37^o+T_2sin53^o = Fg$

$Fg = Mg$
$Fg = 5 * 10$
$Fg = 50$

$T_2 = T_1cos37^o/cos53^o$
$T_2 = 1.327T_1$

$T_1sin37^o+(1.327T_1)sin53^o = 50$
$T_1(sin37^o+1.327*sin53^o) = 50$
$1.662T_1 = 50$
$T_1 = 30.084$

$(30.084)cos37^o = T_2cos53^o$
$24.026 = T_2(0.602)$
$T_2 = 39.9103$

$T_1 = 30.084N$
$T_2 = 39.9103N$

 

[haruspex]

I'm sorry; I misinterpreted what was written on my textbook. Upon further inspection, here is what I have ended up with so far. Do any of these seem right?

For the string attached to the wall, our degree is 37; there is a 90 degree angle formed by the dotted lines in our figure above, and 37 subtracted from 90 gives us 53.

$T_1cos37^o = T_2cos53^o$
$T_1sin37^o+T_2sin53^o = Fg$

$Fg = Mg$
$Fg = 5 * 10$
$Fg = 50$

$T_2 = T_1cos37^o/cos53^o$
$T_2 = 1.327T_1$

$T_1sin37^o+(1.327T_1)sin53^o = 50$
$T_1(sin37^o+1.327*sin53^o) = 50$
$1.662T_1 = 50$
$T_1 = 30.084$

$(30.084)cos37^o = T_2cos53^o$
$24.026 = T_2(0.602)$
$T_2 = 39.9103$

$T_1 = 30.084N$
$T_2 = 39.9103N$

Yes, except you haven't said which is T₁ and which T₂.

 

[Masrat_A]

Oh, I'm sorry. T1 is the ceiling, and T2 is our wall.

 

[haruspex]

Oh, I'm sorry. T1 is the ceiling, and T2 is our wall.

Which would you expect to have the greater tension?

 

[Masrat_A]

I would imagine the ceiling to have greater tension.

 

[haruspex]

I would imagine the ceiling to have greater tension.

So what do you think you might have done wrong?

 

[Masrat_A]

So what do you think you might have done wrong?

Perhaps I've gotten the degrees the other way?

 

[haruspex]

Perhaps I've gotten the degrees the other way?

So let's check your first equation in post #9, the horizontal balance of forces. What is the horizontal component of T₁?

 

[Masrat_A]

Should it have been $T_1 = vcosϴ$ rather than $T_1 = cosϴ$? I apologize; I'm having a very tough time pinpoint precisely what went wrong.

 

[haruspex]

Should it have been $T_1 = vcosϴ$ rather than $T_1 = cosϴ$? I apologize; I'm having a very tough time pinpoint precisely what went wrong.

Neither of those suggestions make sense. A component of T₁ would be T₁ multiplied by the sine or cosine of some angle. Previously you multiplied it by cos(37°). Was that right?

 

[Masrat_A]

No, that was not right. I feel like I'm understanding what's happening here. In our figure, $T_1$, which is attached to the wall, is stated to be $37^o$. Between both strings, there is a $180^o$ angle; therefore, $T_1$, our ceiling, would be $180 - 37 - 53 - 37 = 53^o$.

$T_1cos53^o = T_2cos37^o$
$T_1sin53^o+T_2sin37^o = Fg$

$Fg = Mg$
$Fg = 5 * 10$
$Fg = 50$

$T_2 = T_1cos53^o/cos37^o$
$T_2 = 0.754T_1$

$T_1sin53^o+(0.754T_1)sin37^o = 50$
$T_1(sin53^o+0.754*sin37^o) = 50$
$1.252T_1 = 50$
$T_1 = 39.936$

$(39.936)cos53^o = T_2cos37^o$
$24.034 = T_2(0.799)$
$T_2 = 30.080$

$T_1 = 39.936N$
$T_2 = 30.080N$

 

[haruspex]

That's it!